18
$\begingroup$

This question already has an answer here:

Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$

I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$

$\endgroup$

marked as duplicate by user147263, Marconius, Cameron Williams, BLAZE, user149792 Nov 19 '15 at 4:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

26
$\begingroup$

The identity $k^2-1=(k+1)(k-1)$ shows that $$ \prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{k^2-1}{k^2}=\prod_{k=2}^n\frac{k-1}k\cdot\prod_{k=2}^n\frac{k+1}k=\frac1n\cdot\frac{n+1}2, $$ and the value of limit should follow.

$\endgroup$
  • 3
    $\begingroup$ We can call it a "telescoping product". $\endgroup$ – GEdgar Jan 25 '13 at 18:45
  • $\begingroup$ @GEdgar Absolutely. One probably applies telescopic more often to sums but this fits perfectly products as well. $\endgroup$ – Did Jan 25 '13 at 20:07
9
$\begingroup$

The $r$th term is $\frac{r^2-1}{r^2}=\frac{(r-1)}r\frac{(r+1)}r$

So, the product of $n$ terms is $$\frac{3.1}{2^2}\frac{4.2}{3^2}\frac{5.3}{4^2}\cdots \frac{(n-1)(n-3)}{(n-2)^2}\frac{(n-2)n}{(n-1)^2}\frac{(n-1)(n+1)}{n^2}$$ $$=\frac12\frac32\frac23\frac43\cdots\frac{n-2}{n-1}\frac n{n-1}\frac{n-1}n\frac{n+1}n=\frac12\frac{(n+1)}n$$

as the 1st half of any term is cancelled by the last half of the previous term except for the 1st & the last term.

$\endgroup$
5
$\begingroup$

Your way works nice if you employ the Euler's infinite product for the sine function. Then

$$\lim_{n \to \infty}\sum_{k=2}^{n} \ln\left(1-\frac{1}{k^2}\right)=\lim_{x\to\pi}\ln\left(\frac{\pi^2\sin x}{x(\pi^2-x^2)}\right)=\lim_{y\to0}\ln\left(\frac{\pi^2\sin y}{y(2\pi-y)(\pi-y)}\right)=\ln(1/2)$$ Thus, your product is $1/2$.

$\endgroup$
  • $\begingroup$ To Chris,s sister would you like to explain me Infinite product of sine function and how can i write $\displaystyle \lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)=\lim_{x\to\pi}\ln\left(\frac{\pi^2\sin x}{x(\pi^2-x^2)}\right)$ Thanks $\endgroup$ – juantheron Jan 25 '13 at 18:58
  • $\begingroup$ @juantheron: I simply used Euler's infinite product for the sine function, took log of both sides and then rearranged things to get the expression you see after the first equal sign. Then it remained to take the limit to $\pi$ to get our limit. For more details on infinite product of sine function, see here: ams.org/bookstore/pspdf/gsm-97-prev.pdf $\endgroup$ – user 1357113 Jan 25 '13 at 19:38
  • $\begingroup$ @juantheron: welcome! $\endgroup$ – user 1357113 Jan 25 '13 at 21:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.